Overhead Line Design Exercise Learning Objectives: Designing an Overhead Transmission Line Appreciate the
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Overhead Line Design Exercise Learning Objectives:
Designing an Overhead Transmission Line
Appreciate the importance of relative International Standard
Effective use of standard softwares in Engineering Analysis and Modelling
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Overhead Line Design Exercise: A new double circuit 400kV twin bundled conductor transmission system tower is required. The span length of this line is set to 300m. You are asked to calculate the circuit rating at maximum operating temperature and the tower dimensions required for proper operation given the following information: Table I: Conductor Details Conductor type: AAAC Rubus Diameter: 31.5mm Number of conductors in bundle 2 Weight: 1.686kg/m (16.53N/m) Stringing tension at installation: 34.6kN at Stringing tension at installation: 15 oC Coefficient of linear expansion (εt): 23e‐6 C‐1 0.0678 Ω/km (at 70oC) Equivalent ac resistance at max temperature: Maximum operating temperature (assume 70°C elastic expansion): Insulator String Details Required creepage distance: 12.1 m Available ratio of creepage to length: 3.5 Length of fittings: 0.4m Other Information Ground clearance: 8.1 m (always required) Shielding angle: 20o Wind speed at which clearances must be 10 m/s maintained: Maximum reduction in distance between 2.1 m phase conductors due to vertical movement: Minimum heat loss due to convection: 43.55 W/m 2
Shield Wire
Insulator Length
Top Cross Arm
Creepage length to prevent tracking
Level 1
Insulator Length with fittings Middle Cross Arm Conductor to cross‐arm clearance must be sufficient to prevent flashover
Level 2
Reduced Tower clearance owing to Swing Angle conductor swing
Bottom Cross Arm
Level 3 Conductor to conductor clearance must prevent flashover. Voltages larger than phase to earth
Minimum clearance to ground
Conductor to tower clearance prevent flashover and safety clearance infringement
Bottom Conductor Minimum Sag
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Task 1: Calculation of Bottom Cross‐arm height: Calculating the bottom cross arm height is a procedure that requires: 1. Calculation of sag 2. Calculation of insulator length 3. Knowledge of ground clearance regulation distance (according to Voltage Level) 4. Knowledge of the length of fittings: A. Calculation of Sag Conductor sag is the result of elastic elongation formed when the conductor is installed on the OHL structure. The elastic elongation is reversible and related to conductor elasticity, weight and tension. Using the expressions for sag and conductor length given below, in conjunction with the information tabulated in table I calculate:
Sag at 15oC sag at 70oC.
Conductor length at 15oC and 70oC
S Tc
W Rs 2 (1) 8T
l c Rs
W 2 Rs 3 (2) 24T 2
I.
STc – conductor sag at conductor temperature Tc
II.
lc – conductor length within a span, m
III.
W – Conductor weight N/m
IV.
Rs – span length, m.
V.
T – tensile force, N.
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B. Calculation of insulator length For a suspension tower, it is significant that the length of the insulator and the swing angle define how close it will come to the tower structure. Reducing the insulator lengths to the minimum possible will allow cross‐arm lengths to be reduced and therefore more compact overhead line geometries are likely to be produced. The creepage distance, is basically the path along the surface of an insulator, and is directly proportional to the AC operating voltage. As the insulator, is lengthened, the voltage it can withstand increases. According to IEC 60664:1980 "Insulation Co‐ordination within Low‐Voltage Systems Including Clearances and Creepage Distances for Equipment,", creepage is the shortest path between two conductive parts (or between a conductive part and the bounding surface of the equipment) measured along the surface of the insulation. A proper and adequate creepage distance protects against tracking, a process that produces a partially conducting path of localized deterioration on the surface of an insulating material as a result of the electric discharges on or close to an insulation surface. The degree of tracking required depends on two major factors: the comparative tracking index (CTI) of the material and the degree of pollution in the environment. Used for electrical insulating materials, the CTI provides a numerical value of the voltage that will cause failure by tracking during standard testing. IEC 112 provides a fuller explanation of tracking and CTI. Tracking that damages the insulating material normally occurs because of one or more of the following reasons:
Humidity in the atmosphere.
Presence of contamination.
Corrosive chemicals.
Altitude at which equipment is to be operated.
The shortest distance from end to end over and along the insulator surface is called the leakage path or creepage distance. Therefore the insulator length is calculated (by the use of Table I) as: Length of insulator (li) = Required Creepage distance / Available ratio of creepage to length
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Therefore the Bottom Cross‐arm height can be calculated as follows: Bottom Cross‐arm height = Sag at maximum operating temperature + Length of Insulator + Length of fitting + Ground Clearance Regulation
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Task 2: Calculation of Middle and Upper Cross‐arm Height The height of the overhead line structure is not normally defined by the AC operating voltage but will be based on lightning impulse (LI) voltage for lower voltage overhead lines and the switching impulse (SI) voltage for the transmission voltage systems. These voltages, much greater than the AC system voltage, require a large gap to prevent flashover. Lightning impulse voltages are used to simulate a surge caused by a lightning strike. The usual source of a lightning overvoltage is a lightning strike to an overhead line. This results in the production of a fast‐fronted voltage and travelling waves that propagate along the overhead line. Measurements and experience have shown that lightning overvoltages can be simulated in the laboratory using a short duration impulse of either polarity reaching its peak voltage in 1 to 2s and falling to 50% of its peak value in 50s. Switching impulse voltages are used to simulate a surge caused by a switching operation (e.g. connection / disconnection of an overhead line by a circuit breaker). Switching overvoltages become important when it is necessary to test equipment rated at approximately 300kV and above. Below this level, the lightning overvoltage is more significant. The waveform normally used to represent a switching surge has a 250s rise‐time and a 2000s fall‐time. It is known that the rise and fall times of switching overvoltages do vary widely around these figures but the values are chosen to match up with the waveshape giving the lowest flashover distance in air. Impulse tests, whether lightning or switching, generally involve the application of voltages in the region of 10 times the working voltage at the lower voltage levels. This reduces to voltages in the region of 3 times the working voltage at the higher voltage levels. Alternating voltages are used to assess the ability of insulation to withstand the continuous power system voltage and temporary overvoltages that are produced during power system fault conditions. Tests include the application of a voltage 2 to 3 times the working voltage for a short period in the region of one minute (the withstand or overvoltage test) and the application of a voltage to measure partial discharge activity and/or the radio interference level. Indoor equipment is only tested in a dry condition but the external insulation of outdoor equipment is usually wet‐tested by applying standardised rain to a test object. 7
The voltage for 50% probability of flashover against a gap distance of a rod‐plane gap is determined by the formulas tabulated in table II. Table II:
U 50
Strength of a gap for an AC voltage (peak):
3740 (kV ) 8 1 d
U 50 (380k 150k )d (kV )
Strength of a gap for a lightning impulse voltage:
U 50
Strength of a gap for a switching impulse voltage:
3400k (kV ) 8 1 d
Create a simple computational algorithm in MATLAB to plot the voltage for 50% probability of flashover against a range of gap distances of a rod‐plane gap, for an AC voltage, a lightning impulse voltage and a switching impulse voltage, using the formulas of table II.
Calculate the distance between the phase conductors as determined by the switching impulse voltage by using the formula given below. The U50 for a switching impulse can be calculated by the use of table 2, where d is the length of the insulator. d pp
8 1.5 U 50 3400 1.5 U 50
Calculate the middle cross‐arm height (h2) and the upper cross‐arm height (h3) by the use of formulas below. Assume that drr is the maximum reduction in distance between phase conductors, as given by table I.
Middle Cross‐arm height:
h2 h1 d pp d rr
Upper Cross‐arm height:
h3 h2 d pp d rr 8
Task 3: Calculation of Cross‐arm Width Phase to tower clearances (mainly applicable for lattice towers) usually depend on the swing angle of the overhead line insulator‐conductor system. When the wind blows, a wind force is imposed on the conductor. This force can move the entire conductor system towards the tower. The amount of movement is determined by the ratio of the wind force imposed on the conductor to the weight of the conductor system as shown by the figure below and the accompanying equation. The weight and wind load of the insulator are divided by two to refer them to the extremity of the insulator length.
Figure 1. Swing Angle Calculation For a suspension tower, it is significant that the length of the insulator and the swing angle define how close it will come to the tower structure. Reducing insulator lengths to the minimum possible will allow cross‐arm lengths to be reduced and more compact overhead line geometries to be produced. By neglecting the weight and wind load of insulators the swing angle Φ can be calculated as:
F Fi / 2 tan 1 C W n Wi / 2 C
F tan 1 C WC n
Fc is the force exerted on a conductor owing by the wind and can be calculated as:
Fwind
2
v 2 d l70 n
Where:
ρ = air density ‐1.22 kg/m3
v = wind velocity at which clearances must be maintained.
d =diameter of conductor
l70 = conductor length at 70 oC
n = number of conductors in bundle 9
Wc is the weight of the conductor system (twin bundle arrangement) and can be calculated as:
Wc Weight ( N / m) ConductorL ength (l70 ) n The phase to earth clearance dpe can be calculated with the use of formula given below.U50 is the strength of a gap for a switching impulse voltage.
d pe
8 U 50 3400 U 50
Consequently the maximum reduction in distance between phase conductor and tower is given as:
d r li sin where d r is defined as the maximum reduction distance; li is the insulator length; Φ is the swing angle. Therefore the cross‐arm width (dw) can be calculated as:
d w d pe d r Task 4: Calculation of shield wire height According to IEEE Std 1410‐1997, shield wires are grounded conductors placed above the phase conductors to intercept lightning strokes which would otherwise directly strike the phases. Lightning current is diverted to ground through a pole ground lead. To be effective, the shield wire is grounded at every pole. Lightning‐surge current flowing through the pole ground impedance causes a potential rise, resulting in a large voltage difference between the ground lead and the phase conductors. The voltage difference may cause a back flashover across the insulation from the ground lead to one of the phase conductors. The shielding angle is defined in IEEE Std 1410‐1997 as the angle between the vertical line through the overhead ground wire and a line connecting the overhead ground wire with the shielded conductor. To ensure that all lightning strokes terminate on the shield wire rather than on the phase conductors, a shielding angle (as shown by Figure below) of 45° or less is recommended. This is only valid for lines less than 15 m tall with conductor spacings under 2 m. Taller lines require smaller shielding angles. Refer to IEEE Std 1243‐1997 [B27]
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Figure 2. Definition of shielding angle as defined in IEEE 1410‐1997 For this particular transmission lattice tower the shielding angle is 20 degree (defined in table I), so the shield wire height can be calculated as:
(d pe d r ) h4 h3 l i 0 .4 o tan(20 )
Task 5: Calculation of Circuit Rating The power rating of an overhead line (OHL) is defined by the voltage level of the line and the current rating and is usually expressed in MVA. The voltage level affects the size of the tower of the OHL in order to satisfy the minimum ground clearances is defined at the initial stages of the OHL design. The ampacity is the maximum continuous current that corresponds to the maximum allowable conductor temperature set by the system operators and is called thermal rating of the line. The conductor temperature is affected by the heat produced from the current flowing through it, the thermal properties of the conductors and the surrounding ambient conditions. It is raised primarily due to Ohmic losses (I2R) and secondarily due to solar heating. There are also two cooling mechanisms that take place: convection and conductor radiation. Convection is a major source of heat loss and is a function of air temperature and wind speed. The conductor radiation is affected by the air temperature and conductor material and size. The method to calculate the thermal rating considers that the conductor is in thermal equilibrium at the maximum operating temperature. 11
Consequently, the heat gain of a bare overhead conductor from I2R losses and solar heat balances the heat lost by convection and conductor radiation cooling as shown below: The heat gain equals heat lost: qc qr qs I R 2
qc = convection heat loss W/m
qr =radiated heat loss W/m
qs = solar heat gain W/m
I = conductor current, amperes
RTC = AC conductor resistance at operating temperature, Tc, Ω/m
The solar heat and the radiated heat loss can be neglected in this case. The minimum heat loss due to convection is 43.55 W/m, as tabulated in table I. Therefore the current can be calculated as per the equation below:
I
qc RTC
Due to the twin bundled conductor, the total current rating will be: 2 I .
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